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3 years ago

1. Explain Why is Earth'satmosphere important to lifeon Earth?

1 answer:

8 0

The atmosphere is an important part of what makes Earth livable. It blocks some of the Sun's dangerous rays from reaching Earth. It traps heat, making Earth a comfortable temperature. And the oxygen within our atmosphere is essential for life.

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Match the term with its description. Match Term Definition Chloroplast A) An organelle that contains chlorophyll and in which ph

Otrada [13]

Chloroplast is an organelle that contains clorophyll
Clorophyll is a green pigment present in all plants
Energy is a capacity to do work that can produce physical changes
Glucose is a type of sugar that is produced during Photosynthesis

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3 years ago

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What is another word for the sound barrier?

8090 [49]

Answer:

The Answer is B. Destructive Interference

Explanation:

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3 years ago

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HELP! URGENT Which of the following best states the relationship between erosion and deposition?

MariettaO [177]

The answer is A: When the energy transporting sediments diminishes, the sediments settle in a low-lying area; therefore, deposition always follows erosion

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3 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

How many moles are in 110 grams of nahco3

yuradex [85]

Molar mass NaHCO₃ = 23 + 1 + 12 + 16 x 3 = 84 g/mol

1 mole ---------- 84 g
? mole ---------- 110 g

moles NaHCO₃ = 110 . 1 / 84

moles NaHCO₃ = 110 / 84

= 1.309 moles

hope this helps!

4 0

3 years ago

1. Explain Why is Earth'satmosphere important to lifeon Earth?​

1. Explain Why is Earth'satmosphere important to lifeon Earth?